SMO(J) Mock Paper 1

Duration: Three hours

May 31, 2012

1. The sequence {xn} satisfies x1 =1

2, xn+1 = xn +

x2nn2

. Prove that x2001 < 1001.

2. Let n1, n2, · · · , n1998 be positive integers such that

n21 + n2

2 + · · ·+ n21997 = n2

1998.

Show that at least two of the numbers are even.

3. Consider a convex polygon having n vertices, n ≥ 4. We arbitrarily decompose thepolygon into triangles having all the vertices among the vertices of the polygon,such that no two of the triangles have interior points in common. We paint inblack the triangles that have two sides that are also sides of the polygon, in redif only one side of the triangle is also a side of the polygon and in white thosetriangles that have no sides that are sides of the polygon.

Prove that there are two more black triangles that white ones.

4. Let AL and BK be angle bisectors in the non-isosceles triangle 4ABC (L lies onthe side BC, K lies on the side AC). The perpendicular bisector of BK intersectsthe line AL at point M . Point N lies on the line BK such that LN is parallel toMK. Prove that LN = NA.

5. On a given 2012 × 2012 chessboard, every unit square is filled with one of theletters S,M,O, J. The resulting board is called harmonic if every 2× 2 subsquarecontains all four different letters. How many harmonic boards are there?

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Solutions

1. (CWMO 2001) The sequence {xn} satisfies x1 =1

2, xn+1 = xn +

x2nn2

. Prove that

x2001 < 1001.

Solution. First we have x1 =1

2and x2 =

9

16< 1. If xn <

n

2, then

xn+1 = xn +x2nn2

<n

2+

1

n2× n2

4<

n + 1

2.

Hence by induction we must have xn <n

2. So x2001 < 1000.5 < 1001.

2. (JBMO 1997) Let n1, n2, · · · , n1998 be positive integers such that

n21 + n2

2 + · · ·+ n21997 = n2

1998.

Show that at least two of the numbers are even.

Solution. Firstly, if exactly one of n1, n2, · · · , n1998 is even, then n1998 must notbe even since it is a sum of 1997 odd numbers. If exactly one of x1, x2, · · ·x1997 iseven, then x1998 must be even since it is a sum of one even number and 1996 oddnumbers. Hence there cannot be exactly one even number.

If all numbers are odd, we observe that the remainder when an odd square numberis divided by 8 is always 1. As such,

n21 + ... + n2

1997 ≡ 1997 ≡ 5 ≡ n21998 (mod 8)

which is a contradiction. Hence there must be at least two of the numbers whichare even.

3. (JBMO 2004) Consider a convex polygon having n vertices, n ≥ 4. We arbitrarilydecompose the polygon into triangles having all the vertices among the vertices ofthe polygon, such that no two of the triangles have interior points in common. Wepaint in black the triangles that have two sides that are also sides of the polygon,in red if only one side of the triangle is also a side of the polygon and in whitethose triangles that have no sides that are sides of the polygon.

Prove that there are two more black triangles that white ones.

Solution. Let b be the number of black triangles, r be the number of red triangles,w be the number of white triangles. The sum of interior angles of all the trianglesis equal to the sum of interior angles of the polygon. Hence, we have

b + r + w = n− 2.

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Moreover, since each black triangles share 2 common sides with the polygon whileeach red triangles share 1 common side with the polygon, we have

2b + r = n.

Subtracting the two equations we obtain b = w + 2 and hence there are 2 moreblack triangles than white triangles.

4. (JBMO 2010) Let AL and BK be angle bisectors in the non-isosceles triangle4ABC (L lies on the side BC, K lies on the side AC). The perpendicular bisectorof BK intersects the line AL at point M . Point N lies on the line BK such thatLN is parallel to MK. Prove that LN = NA.

Solution. First, we construct the circumcircle of4AKB and suppose it intersectswith line AL at point M ′. We note that M ′ is the midpoint of the minor arc BK.Hence, M ′ lies above the perpendicular bisector of BK and hence M ′ coincideswith point M .

This shows that ABMK is a cyclic quadrilateral. From there we have ∠LAB =∠BKM = ∠BNL and hence ABNL is a cyclic quadrilateral too. Finally, since∠ABN = ∠LBN , we have LN = NA.

5. (CGMO 2008) On a given 2012× 2012 chessboard, every unit square is filled withone of the letters S,M,O, J. The resulting board is called harmonic if every 2× 2subsquare contains all four different letters. How many harmonic boards are there?

Solution. We consider the problem in the following two cases:

(a) If the first row only contains two letters. Without loss of generality, let usassume that S,M are selected on the first row. In this case, the first row mustbe in the form of SMSM · · ·SM or MSMS · · ·MS. There are 2 possibilitiesfor the next row, which is either OJOJ · · ·OJ or JOJO · · · JO. Similarly,there will be 2 possibilities for each row on the chessboard. The total number

of possibilities in this case is

(4

2

)× 22012 = 6× 22012.

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(b) If the first row contains three or more letters, we note that the entire chess-board is determined just by the first row of the chess board. In this case,we observe that the first column of the chess board only contains 2 letters.Hence, this case is similar to the previous case, just that the chessboard isbeing rotated clockwise by 90◦. We still need to deduct the number of caseswhere both the first row and column only has 2 letters. The total number of

cases is

(4

2

)× 22012 − 4× 3× 2 = 6× 22012 − 24.

Hence, the total number of cases is 12× 22012 − 24.

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